Question #01170

1 Answer
Mrs. Goldcamp · Stefan V. · mason m
Jun 8, 2016

The #"pH"# is about #5.1#

Explanation:

Because ammonium #("NH"_4^+)# is a weak acid, you need to determine the amount of #"H"^+# in the solution by setting an equilibrium problem.

Step 1: find #K_"a"# of ammonium from the #K_"b"# of ammonia (acid / base conjugates).

#K_"a" * K_"b" = 10^-14#

#K_"a"# of #("NH"_4)^(+)=10^-14/(K_text(b))=5.6 xx 10^-10#

Step 2: Set up the dissociation of ammonium equilibrium, and find the amount of #text(H)^+# in solution at equilibrium.

#{: (," ",bb("NH"_4^+),bb->,bb"NH"_3,+,bb("H"^+)), ("initial:",,"0.100 M",,"0 M",,"0 M"), ("change:",,-x,,+x,,+x), ("equilibrium:",,0.1-x,,x,,x) :}#

You can approximate that

#0.1 - x ~~ 0.1#

b/c #K_"a"# is very small relative to #"0.1 M"#.

#K_"a" = (["NH"_3] * ["H"^+])/(["NH"_4^+]) = (x * x) /0.1 = 5.6 xx 10^-10#

#x^2 = 0.1 * 5.6 xx 10^-10#

#x = sqrt(5.6 xx 10^-11) =7.5 xx 10^-6#

#["H"^+] = x = 7.5 xx 10^-6#

Step 3: calculate #"pH"# from the concentration of #"H"^+# found in step 2.

#"pH" = -log["H"^+] = -log(7.5xx10^-6) = 5.1#

Related questions
See all questions in pH calculations
Impact of this question
8098 views around the world
You can reuse this answer
Creative Commons License